Question

Prove it :
$\frac{1 + \cos(x) - \sin(x) }{1 + \cos(x) + \sin(x) } = \sec(x) - \tan(x)$​
Please show working using paper...

Answer 1

Heya!

$\text{First, manipulate the left side.}\\\text{Use the rule:}~a-b=\frac{(a-b)(a-b)}{a+b} = \frac{a^2-b^2}{a+b} \\\text{Use the identity:}~1-sin^2(x)=cos^2(x)\\\frac{1+cos(x)-sin(x)}{1+cos(x)+sin(x)} = \frac{cos(x)+\frac{cos^2(x)}{1+sin(y)}}{1+cos(x)+sin(x)}$

$\text{Second, simplify the numerator}~cos(x)+\frac{cos^2(x)}{1+sin(x)}\\\text{Convert element to a fraction:}~cos(x)=\frac{cos(x)(1+sin(x))}{1+sin(x)}\\\text{Add:}~cos(x)=\frac{cos(x)(1+sin(x))}{1+sin(x)}+\frac{cos^2(x)}{1+sin(x)}\\\text{The denominators are equal so combine:}~\frac{cos(x)(1+sin(x))+cos^2(x)}{1+sin(x)}\\\text{Simplify:} \frac{\frac{cos(x)(sin(x)+1)+cos^2(x)}{1+sin(x)} }{1+cos(x)+sin(x)}\\ \text{Apply fraction rule:}~\frac{cos(x)(1+sin(x))+cos^2(x)}{1+sin(x)}$

$\text{Factor:}~\frac{cos(x)(1+sin(x)+cos(x))}{(1+sin(x))(1+cos(x)+sin(x))} \\\text{Simplify:}~\frac{cos(x)}{1+sin(x)}$

$\text{Third, manipulate the right side.}\\\text{Use the basic trigonometric identity:}~sec(x)=\frac{1}{cos(x)} \\\text{Use the basic trigonometric identity:}~tan(x)=\frac{sin(x)}{cos(x)} \\\text{Put the expression back together:}~\frac{1}{cos(x)}-\frac{sin(x)}{cos(x)}\\\text{Simplify:}~\frac{\frac{cos^2(x)}{1+sin(x)} }{cos(x)}$

$\text{Fourth, simplify.}\\\text{Apply the fraction rule:}~\frac{cos^2(x)}{(1+sin(x))cos(x)} \\\text{Cancel out the common factor:}~\frac{cos(x)}{1+sin(x)}$

Therefore, the expression is TRUE.

Best of Luck!